Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
 A: Evaluate the equation for different $x$:
$$\begin{align}\\
\frac\pi6&\to -a+b+2c&+d&=\frac1{64}\\
\frac\pi4&\to -a-b+\ c&+d&=\frac8{64}\\
\frac\pi3&\to +a-b&+d&=\frac{27}{64}\\
\frac\pi2&\to -a+b-\ c&+d&=\frac{64}{64}\\
\end{align}$$
(after normalization of the coefficient of $d$).
Elimination by the combination $(1)+3\times(3)+2\times(4)$ yields
$$6d=\frac{210}{64}.$$
The evaluation for $x=\dfrac\pi4$ was not even necessary.
A: $$\begin{align} & \sin^7x  \\ & 
=\sin^6x\cdot \sin x \\ &
=\frac{1}{16}\cdot 16\sin^6x\cdot \sin x \\  &
=\frac{1}{16}\cdot (4\sin^3x)^2 \cdot \sin x \\  &
=\frac{1}{16}\cdot (3\sin x-\sin 3x)^2 \cdot \sin x \\ &
=\frac{1}{16}\cdot (9\sin^2 x-6\sin x\sin 3x+\sin^2 3x) \cdot \sin x \\ &
=\frac{1}{16}\cdot [\frac{9}{2}(1-\cos 2x)-3(\cos 2x-\cos 4x)+\frac{1}{2}(1-\cos 6x)] \cdot \sin x \\ &
=\frac{9}{32}(\sin x-\sin x\cdot \cos 2x)-\frac{3}{16}(\sin x\cdot \cos 2x-\sin x\cdot \cos 4x)+\frac{1}{32}(\sin x-\sin x\cdot \cos 6x) \\ &
=\frac{9}{32}[\sin x-\frac{1}{2}(\sin 3x- \sin x)]-\frac{3}{32}[(\sin 3x- \sin x)-(\sin 5x - \sin 3x)]+\frac{1}{32}[\sin x-\frac{1}{2}(\sin 7x-\sin 5x)] \end{align}$$
The coefficient of $\sin x$ in the above expression is $$\color{red}{\frac{9}{32}+\frac{9}{64}+\frac{3}{32}+\frac{1}{32}}=\color{blue}{\frac{35}{64}}$$
A: If you only need to know what $d$ is, then consider $z=\cos\theta+i\sin\theta$, in which case $$\left(z-\frac 1z\right)^7=(2i\sin\theta)^7=...+35z^4\left(-\frac 1z\right)^3+35z^3\left(-\frac 1z\right)^4+...$$
Hence
$$2^7i^7\sin^7\theta=...-35\left(z- \frac 1z\right)...=...-35.2i\sin\theta+...$$
therefore the required $$d=\frac{35}{64}$$
A: This is the Fourier series for $f(x) = \sin^7 x$ on the domain $[0,2 \pi]$.
You have $\int_0^{2 \pi} \sin^7 x \sin x dx = d\int_0^{2 \pi} \sin x \sin x dx$.
Here is the answer:

 This gives ${35 \over 64 } \pi = d \pi$.

To compute the integral,
\begin{eqnarray}
\int_0^{2 \pi} \sin^8 x &=& { 1\over (2i)^8} \int_0^{2 \pi} (e^{it}-e^{-it})^8 dt \\
&=& { 1\over (2i)^8} \sum_{k=0}^8 \binom{8}{k} \int_0^{2 \pi} e^{it(8-2k)} dt \\
&=& { 1\over (2i)^8} \binom{8}{4} \int_0^{2 \pi} 1 dt \\
&=& {35 \over 64} \pi
\end{eqnarray}
A: Interesting answers .Let me add another, just for filing a somewhat more "theoretical" one, using Chebyshev polynomials $U_n(x)$ of the second kind defined by
$$U_0(x)=1\\U_1(x)=2x\\U_{n+1}(x)=2x\space U_n(x)-U_{n-1}(x)$$
and for which one has the formula
$$U_n(\cos x)=\frac{\sin (n+1)x}{\sin x}\qquad (*)$$ 
We use the data (if you don’t want to calculate it)
$$\begin{cases}U_6(x)=64x^6-80x^4+24x^2-1\\U_4(x)=16x^4-12x^2+1\\U_2(x)=4x^2-1\\U_0(x)=1\end{cases}$$
Therefore, using $(*)$ and simplifying  the given identity,we have
$$(\sin x)^6=aU_6(\cos x)+bU_4(\cos x)+cU_2(\cos x)+dU_0(\cos x)$$
It follows, putting  $\cos x= t$,
$$(1-t^2)^3=64at^6+(-80a+16b)t^4+(24a-12b+4c)t^2+(-a-c+b+d)$$
Hence the easy system
 $$\begin{cases}64a=-1\\-80a+16b=3\\24a-12b+4c=-3\\-a+b-c+d=1\end{cases}$$
which is resolved line by line giving in succession 
$$(a,b,c,d)=\left(\frac {-1}{64},\frac {7}{64},\frac{-21}{64},\frac {35}{64}\right)$$
